Toán đại 8

M

muttay04

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M

mikelhpdatke

b)Ta có:

$ a^2-4ab+5b^2+10a-22b+28$
$=[5^2+a^2+(2b)^2-4ab+10a-20b]+(b^2-2b+1)+2$
$=(a-2b+5)^2+(b-1)^2+2$
$\rightarrow $ tìm được cặp nghiệm $(a,b)=(-3,1)$
 
T

thaiha_98

Làm nốt câu 1
Chắc đề là: $\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}$
Giải như sau:
$\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}$
$=\dfrac{a^2}{a^2-(b+c)^2+2bc}+\dfrac{b^2}{b^2-(a+c)^2+2ac}+\dfrac{c^2}{c^2-(a+b)^2+2ab}$
$=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}$
$=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}$
$=\dfrac{a^3+b^3+c^3}{2abc}$ (1)
Ta lại có:
$a+b+c=0$ \Rightarrow $a^3+b^3+c^3=3abc$ (2)
Từ (1) và (2) suy ra:
$\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}=\dfrac{3abc}{2abc}=\dfrac{3}{2}$
Vậy...
 
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